Step 1: Fragmentation at router \(R\).
The original IP datagram size is:
\[
1400 \text{ (TCP data)} + 20 \text{ (IP header)} = 1420 \text{ bytes}
\]
The MTU between \(R\) and \(Q\) is 820 bytes, so fragmentation is required. Each fragment has a 20-byte IP header.
Maximum data per fragment:
\[
820 - 20 = 800 \text{ bytes}
\]
Thus:
- First fragment data = 800 bytes \(\Rightarrow\) fragment size = 820 bytes
- Remaining data = \(1400 - 800 = 600\) bytes \(\Rightarrow\) fragment size = \(600 + 20 = 620\) bytes
Hence, option (A) is correct.
Step 2: Fragment loss handling.
IP is unreliable and routers do not retransmit lost fragments. Therefore, option (B) is incorrect.
Step 3: TCP-level retransmission.
If any fragment is lost, the entire IP datagram is discarded at the receiver. TCP detects loss and retransmits the entire TCP segment from the sender \(P\). Hence, option (C) is correct.
Step 4: TCP header analysis.
Only the first fragment contains the TCP header. The second fragment does not carry TCP port information, so option (D) is incorrect.
Step 5: Conclusion.
The correct statements are (A) and (C).
Final Answer: (A), (C)